Spherical shell rolls down a ramp with friction-time?

A 120 g basketball has a 25.7 cm diameter and may be approximated as a thin spherical
shell.
Starting from rest, how long will it take a basketball to roll without slipping 6.29 m down an incline that makes an angle of 31◦ with the horizontal? The moment of inertia of a thin spherical shell of radius R and mass m is I =2/3mR^2, the acceleration due to gravity is 9.8 m/s^2 , and the coefficient of friction is 0.24. Answer in units of s.

3. The attempt at a solution
I'm having some conceptual difficulties... for rolling, it seems like you would need to track one point on the sphere to find the time it took for that point to travel 6.29 m... but then wouldn't you need theta (in radians)? My approach would be to do that with rotational kinematics equations but then maybe you need to find angular acceleration through torques and whatnot?

In any case, I made a free body diagram, and it seemed like the net torque was cause by the force due to friction only. The net force was the parallel component of gravity (mgsinθ) minus the force due to friction. Through this setup I wanted to find α. I might be overcomplicating it but I didn't want to underestimate the rolling thing... I find that idea very challenging.

I substituted the known values in for the acceleration=angular a * r, Net torque=Iα

Fnet=ma=mgsinθ - F of friction= mgsinθ - μmg
=(120 g)sin31 - (0.24)mg

a/r=α

τ=r*Force of friction=r(0.24)mg
τ=Iα=I(a/r)=r(0.24)mg

Inputted some values and tried to simplify but it got way too complicated--can someone just tell me if my basic setup/idea is right?

If you take x-axis along the ramp and y-axis perpendicular to ramp,
Σfx = Mgsinθ - fr = M*a(cm),....(1)
where fr is the frictional force and a(cm) is the acceleration of center of mass.
Net torque Στ = fr*R = I*α = 2/3*M*R^2*α...(2)
Frictional force fr = μ*M*g*cosθ..(3)
a(cm) = R*α...(4)
Using all these equations, find a(cm)
Then using kinematic equation find required time t.